String could not be parsed as XML php android webservice Error

Question

it is not passing url value from php to xml conversion

Warning: SimpleXMLElement::__construct(): Entity: line 141: parser error : EntityRef: expecting ';' in /home/u395985035/public_html/ddd/ddd.php on line 34

Warning: SimpleXMLElement::__construct(): http://teko.gogo.com/gtw_c?tem_id=844&erw_id=44545 in /home/u395985035/public_html/ddd/ddd.php on line 34

Warning: SimpleXMLElement::__construct(): ^ in /home/u395985035/public_html/ddd/ddd.php on line 34

Fatal error: Uncaught exception 'Exception' with message 'String could not be parsed as XML' in /home/u395985035/public_html/ddd/ddd.php:34 Stack trace: #0 /home/u395985035/public_html/ddd/ddd.php(34): SimpleXMLElement->__construct('\n\t\t

    <?php

  include("config.php"); 

    $sql = "SELECT * FROM table1";
    $q   = mysqli_query($conn,$sql) or die(mysql_error());
    $xml = "<demo1>\n\t\t";

    while($r = mysqli_fetch_array($q))   {
      $xml .= "<t_view>\n\t";
      $xml .= "<l1>".$r['id']."</l1>\n\t\t";  
      $xml .= "<l2>".$r['link1']."</l2>\n\t\t";
      $xml .= "</t_view>\n\t";  
    }

       $xml .= "</demo1>\n\t\t";
    $sxe = new SimpleXMLElement($xml);
    $sxe->asXML("ddd.xml");

 /* I am parsing url link with parameters in $r['link1'] */
?>

I couldn't reproduce this error , it seems there could be problem in your records , I created the same table with 'id' and 'link1' fields , inserted some dummy data and run the codes , it is working fine. — Sachin Vairagi, Sep 28 '16 at 05:19
only simple string of url is working like "https://www.google.co.in/" but "http://teko.gogo.com/gtw_c?tem_id=844&erw_id=44545" this kind of link produce error. — JnT, Sep 28 '16 at 09:04
ok I am able to reproduce it , please use urlencode('http://teko.gogo.com/gtw_c?tem_id=844&erw_id=44545') function before inserting link1 in database, I have tested it and it is working , let me know if have any issue. — Sachin Vairagi, Sep 28 '16 at 13:12

Sachin Vairagi · Answer 1 · 2016-09-30T12:09:41.310

0

Okay don't encode your url just try this , I have tested and it is working.

while($r = mysqli_fetch_array($q))   {
      $xml .= "<t_view>\n\t";
      $xml .= "<l1>".$r['id']."</l1>\n\t\t";  
      $xml .= "<l2>".preg_replace('#&(?=[a-z_0-9]+=)#', '&amp;', $r['link1'])."</l2>\n\t\t";
      $xml .= "</t_view>\n\t";  
    }

edited Sep 30 '16 at 12:09

answered Sep 30 '16 at 11:52

Sachin Vairagi

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String could not be parsed as XML php android webservice Error

1 Answers1